Hi,
when checking the first monad law (left unit) for the IO-monad (and also
for the ST monad):
return a >>= f ≡ f a
I figured out that there is the "distinguishing" context (seq [] True)
which falsifies the law
for a and f defined below
> let a = True
> let f = \x -> (undefined::IO Bool)
> seq (return a >>= f) True
True
> seq (f a) True
*** Exception: Prelude.undefined
Is there a side-condition of the law I missed?
Regards,
David
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